Question: Let $g$ be a function defined for all real numbers except for $x=2$. Also let $g'$, the derivative of $g$, be defined as $g'(x)=\dfrac{x^2}{(x-2)^3}$. On which intervals is $g$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\infty,0)$ and $(0,2)$ (Choice B) B $(-\infty,0)$ and $(2,\infty)$ (Choice C) C $(2,\infty)$ only (Choice D) D $(0,2)$ only (Choice E) E The entire domain of $g$
Explanation: We can analyze the intervals where $g$ is increasing/decreasing by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $g'(x)=\dfrac{x^2}{(x-2)^3}$ and that $g$ is undefined at $x=2$. $g'(x)=0$ for $x=0$. Our critical point is $x=0$, and we should also consider $x=2$. Our points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $(-\infty,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,0)$ $x=-1$ $g'(-1)=-\dfrac{1}{27}<0$ $g$ is decreasing $\searrow$ $(0,2)$ $x=1$ $g'(1)=-4<-1$ $g$ is decreasing $\searrow$ $(2,\infty)$ $x=3$ $g'(3)=9>0$ $g$ is increasing $\nearrow$ In conclusion, $g$ is increasing over the interval $(2,\infty)$ only.